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Operations Scheduling
Krishna Murari
Scheduling
Scheduling is process of setting up operations processing time so that job is completed by the time they are due.
The main objective of scheduling to provide the best service to the customer through efficient use of resources.
Purpose of Scheduling
To help the firm to maximise customer satisfaction
To Minimize service delays To Enable the firm to allocate the
capacity to meet customers’ demands on time.
To reduce cost and maximise profit To create flexibility to accommodate
variation in customer demand
Manufacturing Execution Systems
Continuous – Assembly type – large number of units of homogenous product is produced
Intermittent – production of variety of product one at a time or in batches (custom made as per customer’s requirement)
Combination – neither strictly continuous nor intermittent
Sequence in which waiting jobs are processed is critical to efficiency and effectiveness of the intermittent system
Work Center
A work center is an area in a business in which productive resources are organized and work is completed
Can be a single machine, a group of machines, or an area where a particular type of work is done
Scheduling Methods
Infinite loading : jobs are assigned to work centers without regard to the work center’s capacity as if it’s capacity is infinite. Gantt load charts and visual load profiles and assignment algorithm are used to evaluate loading and assigning the jobs.
Finite loading : A scheduling procedure that assigns jobs into work centers and dermines their starting and completion dates by considering the work center’s capacity. Work center’s capacity is allocated unit by unit by simulating job starting and completion time.
Selection of scheduling method depends on volume of production, capacity of work center and nature of operations
Scheduling Methods
Forward scheduling : Determining the start and finish times for the waiting jobs by assigning them to the earliest available time slots at the work centre. (Used in job shops). Forward scheduling is simple in use and it gets job done in shortest lead time but has excessive inventory as jobs get accumulated at various work centers as they get completed before the availability of next work center.
Backward scheduling : Determining the start and finish times for the waiting jobs by assigning them to the latest available time slots at the work centre to enable to complete each job just when it is due. (Used in assy). It minimizes the inventory since job is not completed till it must go to next work center on its routing.
Scheduling Methods
2 jobs are to be processed on 2 machines. The route sheets is given below. Both jobs should be ready in 8 hours. Develop schedule using forward and backward scheduling.
Job X route Sheet Job Y route sheetRouting sequence
Machine Processing Time Hrs
Routing sequence
Machine Processing Time Hrs
1 1 2 1 1 2
2 2 3 2 2 3
3 1 1
Total 6 5
Scheduling MethodsForward Scheduling
Cu
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in H
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1
2
3
4
5
6
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8
X 1
Y 1
X 2
Y 2
X 3
Scheduling MethodsBackward Scheduling
Job X route Sheet Job Y route sheetRouting sequence
Machine Processing Time Hrs
Routing sequence
Machine Processing Time Hrs
1 1 2 1 1 2
2 2 3 2 2 3
3 1 1
Total 6 5
Scheduling Methods
Backward Scheduling
Cu
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lati
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in H
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1
2
3
4
5
6
7
8
Y 1
X 1
Y 2
X 2
X 3 Y2
Typical Scheduling and Control Functions
Allocating orders, equipment, and personnel
Determining the sequence of order performance
Initiating performance of the scheduled work
Shop-floor control
Work-Center Scheduling Objectives
Meet due dates
Minimize lead time
Minimize setup time or cost
Minimize work-in-process inventory
Maximize machine utilization
Scheduling Activities
Routing : it is specifications of work flow. It explains the sequence of operations and processes to be followed in order to produce a particular product. It defines the what to do and where and how to do. Route sheets give these details.
Loading : It is assigning specific job to each work centre for planned period. For loading capacity limitation of each machine is to be considered.
Dispatching : It is the final act of releasing the job orders to the worker to go ahead with the production process. In this, operation manager releases the job orders as per the planned sequence and then controls the process for effective implementation of schedule.
Dispatching rules called as priority rules are used in scheduling the production activities.
Priority Rules for Job Sequencing 1. First-come, first-served (FCFS) or first-in, first served :
Jobs are processed in the order of their arrival, all jobs are treated as equally important for example – petrol filling
2. Shortest operating time (SOT) or shortest processing time (SPT) : priorities is given based on the shortest processing time of the jobs. This is used when firm wants that maximum no. of jobs should be completed.
3. Earliest due date first (DDate) : The jobs are prioritized according to their earliest due dates. Dispatching is done in such a way that the one with earliest due date is dispatched first, the next earliest next and so on.
Priority Rules for Job Sequencing 4. Slack time remaining (STR) first : In this method, the
operating manager calculate the slack time of each job i.e. the difference between the time remaining in the due date and the processing time required. Job with the shortest slack time is dispatched first.
5. Slack time remaining per operation (STR/OP) : Orders with shortest STR/OP are run first as follows:
Time remaining before due date
Remaining Processing time _
STR/OP =
Number of Remaining operations
Priority Rules for Job Sequencing6. Critical ratio (CR) : This is calculated as the difference
between the due date and the current date by the work remaining. Order with the smallest CR are run first.
7. Last come, first served (LCFS) : This happens frequently by default. As orders arrive, they are placed on the top of the stack and operator using picks up from top and run.
8. Random order or whim : the supervisor or operator selects whichever he feels like running.
9. Longest processing time
10. Start date-due date minus normal lead time.
remaining days of Number
date) Current-date (DueCR
Example of Job Sequencing: First-Come First-Served
Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)
A 4 5 4B 7 10 11C 3 6 14D 1 4 15
Answer: FCFS Schedule
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
What is the FCFS schedule?What is the FCFS schedule?
No, Jobs B, C, and D are going to be late
No, Jobs B, C, and D are going to be late
Do all the jobs get done on time?Do all the jobs get done on time?
Delay A-0, B-1, C-8 and D – 11 days
Example of Job Sequencing: Shortest Operating Time (SOT or SPT)
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Answer: Shortest Operating Time Schedule
Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)
D 1 4 1C 3 6 4A 4 5 8B 7 10 15
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
What is the SOT schedule?What is the SOT schedule?
No, Jobs A and B are going to be late
No, Jobs A and B are going to be late
Do all the jobs get done on time?Do all the jobs get done on time?
Delay D-0, C-0, A-3 and B -5 days
Example of Job Sequencing: Earliest Due Date First (DDate)
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Answer: Earliest Due Date First
Jobs (in order Processing Due Date Flow Timeof arrival) Time (days) (days hence) (days)
D 1 4 1A 4 5 5C 3 6 8B 7 10 15
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
What is the earliest due date first schedule?
What is the earliest due date first schedule?
No, Jobs C and B are going to be late
No, Jobs C and B are going to be late
Do all the jobs get done on time?Do all the jobs get done on time?
Delay D-0, A-0, C-2 and B -5 days
Job Sequencing: Critical Ratio Method
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
What is the CR schedule?What is the CR schedule?
No, but since there is a three-way tie, only the first job or two will be on time
No, but since there is a three-way tie, only the first job or two will be on time
In order to do this schedule the CR’s have be calculated for each job. If we let today be Day 1 and allow a total of 15 days to do the work. The resulting CR’s and order schedule are:CR(A)=(5-4)/15=0.06 (Do this job first)CR(B)=(10-7)/15=0.20 (Do this job second, tied with C and D)CR(C)=(6-3)/15=0.20 (Do this job second, tied with B and D)CR(D)=(4-1)/15=0.20 (Do this job second, tied with B and C)
Do all the jobs get done on time?Do all the jobs get done on time?
remaining days of Number
date) Current-date (DueCR
Job Sequencing:Last-Come First-Served (LCFS)
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Answer: Last-Come First-Served ScheduleJobs (in order Processing Due Date Flow Time
of arrival) Time (days) (days hence) (days)D 1 4 1C 3 6 4B 7 10 11A 4 5 15
No, Jobs B and A are going to be late
No, Jobs B and A are going to be late
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
What is the LCFS schedule?What is the LCFS schedule?Do all the jobs get done on time?Do all the jobs get done on time?
Delay D-0, C-0, B-1 and A -10 days
Slack Time Remaining
No, Jobs D,C and B are going to be late
No, Jobs D,C and B are going to be late
Suppose you have the four jobs to the right arrive for processing on one machine
Suppose you have the four jobs to the right arrive for processing on one machine
What is the STR schedule?What is the STR schedule?
Do all the jobs get done on time?Do all the jobs get done on time?
Delay A-0, D-1, C-2 and B-5 days
Jobs (in order Processing Due Dateof arrival) Time (days) (days hence)
A 4 5B 7 10C 3 6D 1 4
Slack Time for A, B,C and D are (5-4), (10-7), (6-3) and (4-1) , B , C and D have same slack time but D has less processing time so D will be before C &B and sequence will be A,D,C,B
Job sequence
Processing time
Due date Time Flow
A 4 5 4
D 1 4 5
C 3 6 8
B 7 10 15
Scheduling Techniques
Many techniques are used by operations manager. Important techniques are :
Gantt chart
Job sequencing rule
Queuing theory
Critical ratio
Gantt Charts
It was developed by Henry L. Gantt. It is a graphical display of the duration of a set of activities.
Gantt chart: Used as a tool to monitor the progress of work and to view the load on workstations. The chart takes two basic forms: (1) the job or activity
progress chart, and (2) the workstation chart.
The Gantt progress chart graphically displays the current status of each job or activity relative to its scheduled completion date.
The Gantt workstation chart shows the load on the workstations and the nonproductive time.
Gantt Progress ChartGantt Progress Chart
Plymouth
Ford
Pontiac
Job 4/11 6/11 7/11 8/11 9/11 12/115/111/11 2/11 3/11
Current Current date date
Scheduled activity timeActual progress
Start activity
Finish activity
Nonproductive time
Gantt Progress Chart for an Auto Parts Company
Gantt Workstation Chart Gantt Workstation Chart
Gantt Workstation Chart for Hospital Operating Rooms
Johnson’s Job Sequencing Rules
Sequencing of jobs is important for scheduling of jobs. It helps in minimizing the processing time and maximizing the operations efficiency. It also reduces the operations cost over a period of time.
When jobs are processed in multistage, it is necessary to sequence in such a way that idle time is reduced.
Johnson's rules developed by Johnson and Bellman, are utilized to minimized the total time span required for completing the given jobs.
Johnson’s Rule for Two stage Production
If a firm has to perform “n” jobs on two machines A and B in the order AB. Expected time for these jobs are A1, A2 , A3….. An on machine A and B1, B2, B3….. Bn in machine B. Following steps are followed:
1. Identify least processing time in A1, A2,….An and B1, B2….Bn. If there is tie , select either of the processing time.
2. If the smallest processing time is Ar (rth job in machine A then place it at the beginning of the sequence. And if it is Bs (sth job on machine B), then place it at the end of the sequence.
3. If there is tie for the least processing time on machine A, any of the jobs can be placed at the beginning of the sequence. In case of tie at machine B, any of the jobs can be placed at the end of the sequence.
Contd…
Johnson’s Rule for Two stage Production 4. Identify the next least processing time and repeat the above
step. The process is continued till all the jobs are assigned in a sequence. The sequence obtained is called as optimum sequence.
5. Once the time sequence is found, the total elapse time and idle time on machine A and B can be calculated by the following formula.
Total Elapse time = time between start of the first job – end time of last job.
Idle time on machine A = time of completion of last job on machine B in optimum sequence- time of completion of last job on machine A in optimum sequence
Idle time on machine B = Time taken by Machine A in completing first job in sequence - ∑ [ ( time when k th job starts on machine B) – (time when k-1 th Job finishes on B)
Example of Job Sequencing: Johnson’s Rule (Part 1)
Suppose you have the following five jobs with time requirements in two stages of production. What is the job sequence using Johnson’s Rule?
Suppose you have the following five jobs with time requirements in two stages of production. What is the job sequence using Johnson’s Rule?
Time in HoursJobs Stage 1 Stage 2 A 1.50 1.25 B 2.00 3.00 C 2.50 2.00 D 1.00 2.00
Example of Job Sequencing: Johnson’s Rule (Part 2)
First, select the job with the smallest time in either stage.
That is Job D with the smallest time in the first stage. Place that job as early as possible in the unfilled job sequence below.
Drop D out, select the next smallest time (Job A), and place it 4th in the job sequence as it is at stage 2
Drop A out, select the next smallest time. There is a tie in two stages for two different jobs. In this case, place the job with the smallest time in the first stage as early as possible in the unfilled job sequence. Then place the job with the smallest time in the second stage as late as possible in the unfilled sequence.
Job Sequence 1 2 3 4
Job Assigned D A B C
Time in HoursJobs Stage 1 Stage 2 A 1.50 1.25 B 2.00 3.00 C 2.50 2.00 D 1.00 2.00
Problem : A firm produces six type of shoes pairs and manufacturing of shoes pairs require processing on two machines A and B. The processing time for each shoe pair on both machines (in Hrs) are given below:
Shoe pair type
1 2 3 4 5 6
Machine A
35 105 55 25 95 105
Machine B
75 100 95 65 35 20
Determine an optimum sequence so that the total elapse time is minimum. Also calculate idle time on machine A and Machine B
Solution: stage:1 Least processing time is 20 for job 6 at machine B hence it will come at the end of sequence i.e. at 6th place
Shoe pair type
6
Machine A
105
Machine B
20
Solution: stage 2 Next least process time is 25 hrs for shoe pair type 4 on machine A hence it will be placed at the beginning of sequence
Shoe pair type
4 6
Machine A
25 105
Machine B
60 20
Solution: stage: 3 Next least process time is 30 hrs for shoe pair type 1 on machine A and 30 hrs for type 5 pair on machine B so type 1 is scheduled nest to type 4 on machine A and type 5 shoe pair before type 6 pairs shown below
Shoe pair type
4 1 5 6
Machine A
25 35 95 105
Machine B
65 75 35 20
Solution: stage: 4 Next least process time is 55 hrs for shoe pair type 3 on machine , hence it will be sequenced after shoe pair 1
Shoe pair type
4 1 3 5 6
Machine A
25 35 55 95 105
Machine B
65 75 95 35 20
Solution: stage:5 last left our pair 2 will be sequenced after pair 3
Shoe pair type
4 1 3 2 5 6
Machine A
25 35 55 105 95 105
Machine B
65 75 95 100 35 20
Solution: stage:6Now we develop the schedule of each job in optimum sequence as shown below
Job sequence
Machine A Machine B
Time in Processing time
Time out
Time in Processing time
Time out
4 0 25 25 25 65 90
1 25 35 60 90 75 165
3 60 55 115 165 95 260
2 115 105 220 260 100 360
5 220 95 315 360 35 395
6 315 105 420 420 20 440
Note : Time- in for machine will be time-out of the job from machine A or time-out of previous job from machine B which ever is bigger
Solution: stage:7 We can calculate elapse time and idle time as follows
Total elapse time : 440 hours
Idle time for machine A = 440-420 = 20 hrs
Idle time for machine B = Time taken by machine A to complete first job 25 hrs + [ (90-90)+(165-165) + (260-260) +(360-360) + (420-395)] = 50 hrs.
Johnson’s Rule for Three Stage Production
If the expected processing time for n-jobs on 3 machines A, B and C are A1,A2,A3…..An; B1, B2, B3….Bn and C1,C2,C3…..Cn
Stage -1 : check that Johnson rule can be used:
Any one of the following conditions should be satisfied
i) The smallest processing time on machine A should be greater than or equal to the largest processing time on machine B
ii) The smallest processing time on machine C should be greater than or equal to the largest processing time of machine B
Contd..
Johnson’s Rule for Three Stage Production
Stage -2 : Assume two fictitious machines G and H
Stage -3 : Find out corresponding time for these machines by they following formula
Gi = Ai +Bi
Hi = Bi + Ci
Stage -3 : solve the problem like 2 machines and n jobs with order of GH.
The resulting optimum sequence will also become the optimum sequence for the problem of 3 machines and n jobs.
Example : Johnson’s Rule for Three Stage Production
A firm is involved in five types of jobs, each must process in 3 machines A, B, C in the order ABC. The processing time (in hours) for each job on the three machines is given below
Job Processing time
Ai Bi Ci
1 18 10 8
2 19 12 18
3 12 5 16
4 16 6 14
5 21 9 10
Example : Three Stage ProductionFirst we check that condition to apply Johnson’s rule is satisfied. Minimum processing time at machine A is 12 hrs. and maximum processing time at B is 12hrs , it satisfied the condition , hence the rule can be used. Now we use Gi=Ai+Bi andHi=Bi+Ci, the table will be as given below
Job Processing time
Gi = Ai+Bi Hi=Bi+Ci
1 28 18
2 31 30
3 17 21
4 22 20
5 30 19
Example : Three Stage ProductionThe optimum sequence will be as follows
Minimum processing time of job 3 is 17 hrs in G hence it will be first in sequence i.e. job 3 will be first in sequence. Now, in remaining
Jobs, job 1 in machine H is lowest processing time 18 hrs, hence it will be the last in sequence.
Now , in left out jobs, job 5 has lowest processing time in machine H, hence, it will come last but one job i.e. before job 1.
Job 4 has the lowest time in machine H so it will be before the job 5 and job 2 will be in left out slot i.e. after job 3.
Hence sequence will be 3,2,4,5,1. The schedule will be as given in table in next slide.
Example : Three Stage Production
Job sequence
Machine A Machine B Machine C
Time in
Processing time
Time Out
Time in
Processing time
Time Out
Time in
Processing time
Time Out
3 0 12 12 12 5 17 17 16 33
2 12 19 31 31 12 43 43 18 61
4 31 16 47 47 6 53 61 14 75
5 47 21 68 68 9 77 77 10 87
1 68 18 86 86 10 96 96 8 104
Example : Three Stage Production
Total operation time with optimum sequence is 104 hrs
Idle time on machine A = 104 -86 =14 hrs.
Idle time on machine B= 12 + [ (31-17)+ (47-43)+ (68-53) + (77-86) + (104 -96) ]= 62 hrs.
Idle time on machine C = 17 +[ (43-33)+ (61-61)+ (77-75) +(96-87) = 38 hrs.
Johnson’s Rule for n jobs and m-machines
If n-jobs 1,2,3…,n on M machines A1,A2,A3,….Am are to ebsceduled. The processing time of job 1 on m machines are A11, A12, A13…..A1m ; processing time for job 2 are A21, A22, A23…..A2m and processing time for job n are An1, An2, An3….Anm.
Stage -1 : check that Johnson rule can be used:
Any one of the following conditions should be satisfied
i) The smallest processing time on machine A1 should be greater than or equal to the largest processing time on machines A2, A3, A4…. Am-1
ii) The smallest processing time on machine Am should be greater than or equal to the largest processing time of machines A2, A3, A4…. Am-1
Contd..
Johnson’s Rule for n jobs on m machines
Stage -2 : Assume two fictitious machines G and H
Stage -3 : Find out corresponding time for these machines by they following formula
Gi = Ai 1+Ai2 + Ai3…..+Aim-1
Hi = Ai2 + Ai3…..+Aim-1 +Aim
Stage -3 : solve the problem like 2 machines and n jobs with order of GH.
The resulting optimum sequence will also become the optimum sequence for the problem of m machines and n jobs.
A firm is involved in 4 types of jobs, each must process in 4 machines P,Q,R and S in the order PQRS. The processing time (in hours) for each job on the 4 machines is given below
Job Processing time
P Q R S
A 10 7 5 9
B 9 6 4 7
C 8 4 2 6
D 12 8 3 9
Johnson’s Rule for n jobs on m machines
Determine the optimum sequence , elapse time and idle time on each machine
First we check that condition to apply Johnson’s rule is satisfied. Minimum processing time at machine P is 8 hrs. and maximum processing time at Q and R are 8 and 5 , it satisfied the condition , hence the rule can be used. Now we use Gi=A1+A2+A3 A2+A3+A4 the table will be as given below
Job Processing time
Gi = A1+A2+A3 Hi=A2+A3+A4
A 22 21
B 19 17
C 14 12
D 23 20
Johnson’s Rule for n jobs on m machines
The optimum sequence will be as follows
Minimum processing time of job C is 12 hrs in H hence it will be last in sequence.
Now in remaining Jobs, job B in machine H is lowest processing time 17 hrs, hence it will be the last but one in sequence.
Now , in left out jobs, job D has lowest processing time in machine H hence, it will come before job B
Hence sequence will be A,D,B and C. The schedule will be as given in table in next slide.
Johnson’s Rule for n jobs on m machines
Job Machine
P Q R S
A 0-10 10-17 17-22 22-31
D 10-22 22-30 30-33 33-42
B 22-31 31-37 37-41 42-49
C 31-39 39-43 43-45 49-55
Johnson’s Rule for n jobs on m machines
Total operation time with optimum sequence is 55 hrs
Idle time on machine P = 55-39 =16 hrs.
Idle time on machine Q = 10+(22-17)+(31-30)+(39-37)+(55-43) = 30 hrs.
Idle time on machine R = 17+(30-22)+(37-33)+(43-41)+(55-45)= 41 hrs
Idle time on machine S =22+(33-31)+(42-42)+(49-49)=24hrs
Johnson’s Rule for n jobs on m machines
Personnel Scheduling in Services
Issues :
Customer’s direct interaction
High variability in demand
Two operations:
Front of the house
Back of the house
Personnel Scheduling in Services
Scheduling consecutive days off – 5 days week working
Scheduling daily work times
Scheduling hourly work times
Personnel Scheduling in Services
Methods to meet customer demand under limited availability of service facilities
Appointment system – customer arrival time can be controlled
Reservation system – like hotel rooms
Strategic product pricing – to adjust the shift in demand like higher price at peak hours – electricity at high price summer